Find indexes of sequence in list in python

Question

I am quite new and I hope it's not too obvious, but I just can't seem to find a short and precise answer to the following problem.

I have two lists:

a = [2,3,5,2,5,6,7,2]
b = [2,5,6]

I would like to find when all the indexes of the second list (b) are in the first list (a), so that I get something like this:

indexes of b in a: 3, 4, 5 or b = a[3:6]

Answer 1

With a list comprehension:

>>> [(i, i+len(b)) for i in range(len(a)) if a[i:i+len(b)] == b]
[(3, 6)]

Or with a for-loop:

>>> indexes = []
>>> for i in range(len(a)):
...    if a[i:i+len(b)] == b:
...        indexes.append((i, i+len(b)))
... 
>>> indexes
[(3, 6)]

Answer 2

This should do what you are asking:

a = [2,3,5,2,5,6,7,2]
b = [2,5,6]
for i in range(len(a)):
    if a[i] in b:
        print i

Of course you should format the print statement to your own liking. Or don't print and save result in, say, another list.

Good luck!

Answer 3

Also, for efficiency, you can use KMP algorithm that is used in string matching (from here):

def KMPSearch(pat, txt): 
    M = len(pat) 
    N = len(txt) 

    # create lps[] that will hold the longest prefix suffix  
    # values for pattern 
    lps = [0]*M 
    j = 0 # index for pat[] 

    # Preprocess the pattern (calculate lps[] array) 
    computeLPSArray(pat, M, lps) 

    i = 0 # index for txt[] 
    while i < N: 
        if pat[j] == txt[i]: 
            i += 1
            j += 1

        if j == M: 
            print("Found pattern at index " + str(i-j))
            j = lps[j-1] 

        # mismatch after j matches 
        elif i < N and pat[j] != txt[i]: 
            # Do not match lps[0..lps[j-1]] characters, 
            # they will match anyway 
            if j != 0: 
                j = lps[j-1] 
            else: 
                i += 1

def computeLPSArray(pat, M, lps): 
    len = 0 # length of the previous longest prefix suffix 

    lps[0] # lps[0] is always 0 
    i = 1

    # the loop calculates lps[i] for i = 1 to M-1 
    while i < M: 
        if pat[i]== pat[len]: 
            len += 1
            lps[i] = len
            i += 1
        else: 
            # This is tricky. Consider the example. 
            # AAACAAAA and i = 7. The idea is similar  
            # to search step. 
            if len != 0: 
                len = lps[len-1] 

                # Also, note that we do not increment i here 
            else: 
                lps[i] = 0
                i += 1

a = [2,3,5,2,5,6,7,2]
b = [2,5,6]
KMPSearch(b, a) 

This find the first index of the b in a. Hence, the range is the result of the search and its plus to the length of b.